How can I find the points on the curve y = x^4 − 10x^2 + 1 where the tangent line is horizontal?

1 Answer
Feb 26, 2018

Take the first derivative, set it equal to 0, solve for x, plug back the solutions for x into the original function. The tangent line is then shown to be horizontal at (0,1), (sqrt(5),-24), (-sqrt(5),-24).

Explanation:

The tangent line is horizontal where the derivative is equal to zero.

Graphically interpreted, the derivative at a point on a curve is the slope at that point. We draw tangent lines at certain points to represent the slope at certain points. A horizontal line, y=c where c is a constant value, has a slope of zero, so wherever a horizontal tangent line exists, the derivative must be zero.

Take the first derivative with the Power Rule, set y'=0, and solve for x:

y'=4x^3-(2)(10)x=4x^3-20x

4x^3-20x=0

4x(x^2-5)=0

x=0, +-sqrt(5)

So, since y'=0 at x=0, x=+-sqrt(5), the tangent line is horizontal at those points. To find the y-values at these points, take y(0), y(sqrt(5)),y(-sqrt(5)).

y(0)=0^4-10(0^2)+1=1

The tangent line is horizontal at (0,1).

y(sqrt(5))=sqrt(5)^4-10sqrt(5)^2+1=25-50+1=-24

The tangent line is horizontal at (sqrt(5),-24).

y(-sqrt(5))=(-sqrt(5)^4)-10(-sqrt(5)^2)+1=25-50+1=-24

The tangent line is horizontal at (-sqrt(5),-24).